Zero products of Toeplitz operators on Reinhardt domains
aa r X i v : . [ m a t h . C V ] S e p ZERO PRODUCTS OF TOEPLITZ OPERATORS ON REINHARDT DOMAINS ˇZELJKO ˇCU ˇCKOVI ´C, ZHENGHUI HUO, AND S ¨ONMEZ S¸ AHUTO ˘GLUA
BSTRACT . Let Ω be a bounded Reinhardt domain in C n and φ , . . . , φ m be finite sums of boundedquasi-homogeneous functions. We show that if the product of Toeplitz operators T φ m · · · T φ = Ω , then φ j = j .
1. I
NTRODUCTION
Algebraic properties of Toeplitz operators acting on Bergman spaces have been intensivelystudied for the past 30 years. These problems attracted the interests of many operator theorists,partly because they are easy to state but quite difficult to solve. The same problems on theclassical Hardy space of the unit disk D were solved by Brown and Halmos in their famouspaper [BH64]. A problem that is particularly appealing is the zero product problem. Brownand Halmos showed that if the product of two Toeplitz operators T f T g =
0, then either f = g =
0. In that case, we say that the zero product problem has a trivial solution. Thequestion about the zero product of finitely many Toeplitz operators acting on the Hardy spacewas solved much later by Aleman and Vukoti´c [AV09] where they showed the problem alsohas only a trivial solution.For the Bergman space on D , the zero product problem is still open. In their two papers,Ahern and ˇCuˇckovi´c showed that T f T g = f and g are bounded harmonic functions on D [A ˇC01] or if f and g are bounded radial functions[A ˇC04]. The first result was extended by Choe, Lee, Nam and Zheng [CLNZ07] to the case ofthe polydisk provided the symbols are pluriharmonic. Later Choe and Koo [CK06] obtainedthe same conclusion for the unit ball in C n , where the symbols are bounded harmonic func-tions that have continuous extensions to some open set of the boundary. The second resultfrom [A ˇC04] was extended to the unit ball by Dong and Zhou, under the assumption that thesymbols are (separately) quasi-homogeneous [DZ11].So far, all the known results for bounded domains show that the zero product problem has atrivial solution. At this point we would like to mention two works that study the same problemfor unbounded domains. In an interesting paper by C¸ elik and Zeytuncu [C¸ Z16], the authorsconstruct an unbounded domain in C n and a Toeplitz operator that is nilpotent. That is, the zero Date : September 7, 2020.2010
Mathematics Subject Classification.
Primary 47B35; Secondary 32A36.
Key words and phrases.
Toeplitz operator, Reinhardt domain, Bergman space, quasi-homogeneous. product problem has non-trivial solution. Similarly, Bauer and Le [BL11] produced a curiousexample of three nonzero Toeplitz operators on the Fock space whose product is equal to zero.It is worth noting that in both examples of [C¸ Z16] and [BL11], Toeplitz operators have boundedquasi-homogeneous symbols.In this paper we study zero products of finitely many Toeplitz operators acting on the Bergmanspace of bounded Reinhardt domains in C n , with the assumption that the symbols are finitesums of quasi-homogeneous functions. Our results point again in the direction of the trivialsolution. But as opposed to the unit ball and the polydisk, working on more general Rein-hardt domains brings new technical difficulties. Furthermore, additional difficulties come fromworking with finitely many products of Toeplitz operators and choosing the symbols that arefinite sums of quasi-homogeneous functions. We hope that our techniques will bring morelight to the study of algebraic properties of Toeplitz operators on more general domains in C n ,beyond the unit ball and polydisk.Our main result is the following theorem. Theorem 1.
Let Ω be a bounded Reinhardt domain in C n and φ , . . . , φ m be finite sums of boundedquasi-homogeneous functions. Assume that T φ m · · · T φ = on A ( Ω ) . Then φ j = for some j. In particular, our result shows that, the type of examples in [BL11] and [C¸ Z16] cannot happenon bounded domains. In addition, we obtain the following corollary.
Corollary 1.
Let Ω be a bounded Reinhardt domain in C n and φ , . . . , φ m − be bounded quasi-homogeneoussymbols on Ω and let φ m ∈ L ∞ ( Ω ) . Assume that T φ m · · · T φ = on A ( Ω ) . Then φ j = for some j.
2. P
RELIMINARIES
We start this section by some basic definitions. Let Ω be a domain in C n . The space of squareintegrable holomorphic functions on Ω , the Bergman space on Ω , is denoted by A ( Ω ) . Since A ( Ω ) is a closed subspace of L ( Ω ) there exists an orthogonal projection P : L ( Ω ) → A ( Ω ) ,called the Bergman projection of Ω . The Toeplitz operator T φ , with symbol φ ∈ L ∞ ( Ω ) , isdefined as T φ f = P ( φ f ) for all f ∈ A ( Ω ) .In this paper we are focusing on bounded Reinhardt domains and products of Toeplitzoperators with symbols that are finite sums of quasi-homogeneous symbols. So we will de-fine these notions next. A domain Ω is called Reinhardt if ( e i θ z , . . . , e i θ n z n ) ∈ Ω whenever z = ( z , . . . , z n ) ∈ Ω and θ j ∈ R for all j . A function φ is called (separately) quasi-homogeneousif there exists f : [ ∞ ) n → C , ( k , . . . , k n ) ∈ Z n such that φ ( r e i θ , . . . , r n e i θ n ) = f ( r , . . . , r n ) e i ( k θ + ··· + k n θ n ) .We note that such functions are called separately quasihomogeneous in [DZ11]. Next we intro-duce a condition for a set of multi-indices motivated by [DZ11]. ERO PRODUCTS OF TOEPLITZ OPERATORS ON REINHARDT DOMAINS 3
Definition . Let N = N ∪ { } and E be a subset of N n . Let π j be the projection of the multi-indices in N n onto the j th coordinate: π j ( a , . . . , a n ) = a j . We use E ( a , . . . , a j ) to denote thefiber in E with its first j components being a , . . . , a j , respectively. That is, E ( a , . . . , a j ) = { a ∈ E : π k ( a ) = a k for k =
1, 2, . . . , j } .(i). We say the fiber E ( a , . . . , a j ) is thick if π j + ( E ( a , . . . , a j )) ∩ N = ∅ and ∑ k ∈ π j + ( E ( a ,..., a j )) ∩ N k = ∞ .We say E ( a , . . . , a j ) is a thin fiber if it is not thick .(ii). We say E satisfies condition (I) if E contains a subset b E ⊂ N n satisfying the followingconditions:(1) the sum ∑ k ∈ π ( b E ) k = ∞ ,(2) for 1 ≤ j < n , and any j -tuple ( a , . . . , a j ) ∈ N j , if its corresponding fiber in b E b E ( a , . . . , a j ) : = { a ∈ b E : π k ( a ) = a k for k =
1, 2, . . . , j } is nonempty, then b E ( a , . . . , a j ) is thick. Lemma 1.
Let f be a bounded holomorphic function on the product of right half planes H n + = { z ∈ C n : Re ( z j ) > j =
1, . . . , n } . If the set E = { α ∈ N n : f ( α ) = } satisfies condition (I), then f vanishes identically on H n + . Proof.
Since E satisfies condition (I) there exists a subset b E of E ∩ N n such that for any fixedmulti-index ( a , . . . , a n − ) with b E ( a , . . . , a n − ) = ∅ , there exists a sequence { a ( l ) n } such that ( a , . . . , a n − , a ( l ) n ) ∈ b E for each l and ∑ ∞ l = a ( l ) n = ∞ . Then f ( a , . . . , a n − , z n ) ≡ H + for any fixed multi-index ( a , . . . , a n − ) with b E ( a , . . . , a n − ) = ∅ (see [Rem98, Page 102] and[DZ11, Theorem 2.3]). Since b E ( a , . . . , a n − ) ⊃ b E ( a , . . . , a n − , a n − ) = ∅ ,there also exists a sequence { a ( l ) n − } such that b E ( a , . . . , a n − , a ( l ) n − ) = ∅ and ∑ ∞ l = a ( l ) n − = ∞ .Thus for every fixed z n ∈ H + and every ( a , . . . , a n − ) with b E ( a , . . . , a n − ) = ∅ we have f ( a , . . . , a n − , z n − , z n ) ≡ H + . Repeating this process yields f ( z ) ≡ H n + . (cid:3) Lemma 2.
Let Z and Z be subsets of N n such that Z ∪ Z satisfies condition (I). Then Z or Z satisfy condition (I). Proof.
Let M = Z ∪ Z . Since M satisfies condition (I), there exists b M ⊂ ( Z ∪ Z ) ∩ N n satisfying (ii) in Definition 1. Assume that Z does not satisfy condition (I). Then we apply thefollowing process to Z :i. Set E n = Z . ˇZELJKO ˇCU ˇCKOVI ´C, ZHENGHUI HUO, AND S ¨ONMEZ S¸AHUTO ˘GLU ii. Define E n − to be the set obtained by deleting all thin fibers of the form E n ( a , . . . , a n − ) from E n , i.e. E n − = { ( a , . . . , a n ) ∈ E n : E n ( a , . . . , a n − ) is thick } .iii. For 1 ≤ j ≤ n −
2, define E j to be the set obtained by deleting all thin fibers of the form E j + ( a , . . . , a j ) from E j + , i.e. E j = { ( a , . . . , a n ) ∈ E j + : E j + ( a , . . . , a j ) is thick } .iv. If E = ∅ , then we set E = ∅ if ∑ j ∈ π ( E ) j < ∞ , E otherwise.By running the above process, all the thin fibers will be deleted from Z . That is, in step ii.all the remaining non-empty fibers E n − ( a , . . . , a n − ) will be thick. In step iii. we remove (twodimensional) thin (in the ( n − ) -st component) fibers of the form E n − ( a , . . . , a n − ) to get theset E n − . Therefore, if E n − ( a , . . . , a n − ) is non-empty then it is thick and E n − ( a , . . . , a n − ) = E n − ( a , . . . , a n − ) .When deleting the two dimensional fibers from E n − to obtain E n − , a one dimensional fiber E n − ( a , . . . , a n − ) in E n − would either be untouched or entirely removed. In other words, ifthe (one dimensional) fiber E n − ( a , . . . , a n − ) is non-empty then it is thick and E n − ( a , . . . , a n − ) = E n − ( a , . . . , a n − ) .Therefore, the non-empty fibers E n − ( a , . . . , a k ) are thick for n − ≤ k ≤ n −
1. Arguinginductively in step iii., we conclude that the non-empty fibers E j ( a , . . . , a k ) are thick for 1 ≤ j ≤ k ≤ n −
1. It is possible that E j becomes the empty set for some j ≥
1. Nevertheless, E iseither an empty set, or a set with all fibers in it being thick and satisfy ∑ j ∈ π ( E ) j = ∞ . Since Z does not satisfy condition (I), E has to be an empty set. Thus Z = ∪ nj = ( E j \ E j − ) . Set F j = E j \ E j − for j =
1, . . . , n . Then we have the following properties for F j :i. Z = ∪ nj = F j .ii. F = E where E either satisfies ∑ j ∈ π ( E ) j < ∞ or it is empty.iii. For 2 ≤ j ≤ n , F j consists of all thin fibers in E j of the form E j ( a , . . . , a j − ) , i.e. F j = E j \ E j − = { ( a , . . . , a n ) ∈ E j : E j ( a , . . . , a j − ) is thin } for 2 ≤ j ≤ n .Note that if a fiber b M ( a , . . . , a n − ) of b M is thick, then b M ( a , . . . , a n − ) \ F n ( a , . . . , a n − ) is stillthick since F n ( a , . . . , a n − ) is thin. Hence deleting F n from b M does not affect the “thickness”of b M ( a , . . . , a n − ) \ F n ( a , . . . , a n − ) , fibers with their first ( n − ) components fixed. More-over, all non-empty fibers of the form b M ( a , . . . , a n − ) \ F n ( a , . . . , a n − ) are still thick, becausea thick portion of the fiber survives when we remove a thin fiber. To be more precise, if b M ( a , . . . , a n − ) is non-empty then it is thick. Furthermore, since F n ( a , . . . , a n − ) is thin, the ERO PRODUCTS OF TOEPLITZ OPERATORS ON REINHARDT DOMAINS 5 set b M ( a , . . . , a n − ) \ F n ( a , . . . , a n − ) is non-empty and thick. That is, π n − ( b M ( a , . . . , a n − )) = π n − ( b M ( a , . . . , a n − ) \ F n ) .Thus the thickness of the fiber corresponding to the ( n − ) -tuple ( a , . . . , a n − ) stays the same.Similarly the thickness of fibers with their first j components fixed are not affected for j ≤ n − b M \ F n is still a set satisfying (ii) in Definition 1.Now we turn to show that the set ( b M \ F n ) \ F n − contains a subset satisfying (ii) of Definition1. The same argument as in the previous paragraph yields that, for j ≤ n −
2, none of thefibers with their first j components fixed are entirely deleted from b M \ F n and hence all fibersin ( b M \ F n ) \ F n − with their first j components fixed are thick. For each ( n − ) -tuple a ′ =( a , . . . , a n − ) such that ( b M \ F n )( a ′ ) \ F n − ( a ′ ) is nonempty, we have ( b M \ F n )( a ′ ) is thick and F n − ( a ′ ) is thin. Thus ∑ j ∈ π n − (( b M \ F n )( a ′ ) \ F n − ( a ′ )) j = ∞ .Moreover, for each j ∈ π n − (( b M \ F n )( a ′ ) \ F n − ( a ′ )) , the fiber (( b M \ F n ) \ F n − )( a ′ , j ) is thicksince F n − ( a ′ , j ) is empty. Therefore the set b M n − : = { ( a , . . . , a n ) ∈ ( b M \ F n ) \ F n − : F n − ( a , . . . , a n − ) = ∅ } is a subset of ( b M \ F n ) \ F n − satisfying (ii) of Definition 1 which implies that ( b M \ F n ) \ F n − satis-fies condition (I).Similarly, we can show that (( b M \ F n ) \ F n − ) \ F n − , . . . , up to the set b M \ ( ∪ nl = F l ) all satisfycondition (I). Hence b M \ Z satisfies condition (I). It follows from the containment Z ⊇ b M \ Z that condition (I) holds true for the set Z . (cid:3) As a direct consequence of Lemma 1 we have the following corollary.
Corollary 2.
Let Z , . . . , Z m be subsets of N n such that ∪ mj = Z j satisfies condition (I). Then Z j satisfiescondition (I) for some ≤ j ≤ m.
3. P
ROOF OF C OROLLARY AND T HEOREM
Proposition 1.
Let Ω be a bounded Reinhardt domain in C n and φ , . . . , φ m be bounded quasi-homogeneoussymbols on Ω . Assume that T φ m · · · T φ = on A ( Ω ) . Then φ j = for some j. Proof.
Let Ω + and e Ω + denote the subsets in R n defined by Ω + = { ( | z | , . . . , | z n | ) ∈ R n : z = ( z , . . . , z n ) ∈ Ω } ; e Ω + = { ( x , . . . , x n ) ∈ R n : ( x , . . . , x n ) ∈ Ω + } . ˇZELJKO ˇCU ˇCKOVI ´C, ZHENGHUI HUO, AND S ¨ONMEZ S¸AHUTO ˘GLU Since φ , . . . , φ m are quasi-homogeneous, for z = ( r e i θ , . . . , r n e i θ n ) ∈ Ω and each j we set φ j ( z ) = f j ( r ) e i k j · θ where r = ( r , . . . , r n ) , θ = ( θ , . . . , θ n ) , and k j ∈ Z n . The Bergman kernelfunction K Ω has the expression K Ω ( z , w ) = ∑ α ∈ Z n c α z α ¯ w α where c α = k z α k = ∞ k z α k − otherwise. A polar coordinates computation yields for k ∈ N n that T φ z k = Z Ω ∑ α ∈ Z n c α z α ¯ w α φ ( w ) w k dV ( w )= ( π ) n c k + k z k + k Z Ω + f ( r ) r k + k + dV ( r )= π n c k + k z k + k Z e Ω + f ( √ t ) t k +( k /2 ) dV ( t ) ,(1)where √ t = ( √ t , . . . , √ t n ) . Set g ( t ) = f ( √ t ) t k /2 and d ( k ) = π n c k + k . Then (1) becomes T φ z k = d ( k ) z k + k Z e Ω + g ( t ) t k dV ( t ) .Similarly, T φ z k + k = Z Ω ∑ α ∈ Z n c α z α ¯ w α φ ( w ) w k + k dV ( w )= π n c k + k + k z k + k + k Z e Ω + f ( √ t ) t k + k +( k /2 ) dV ( t ) .(2)We also set g ( t ) = f ( √ t ) t k +( k /2 ) and d ( k ) = π n c k + k + k . Then T φ T φ z k = d ( k ) d ( k ) z k + k + k Z e Ω + g ( t ) t k dV ( t ) Z e Ω + g ( t ) t k dV ( t ) .Repeating this process up to T φ m , we obtain T φ m · · · T φ z k = z k m ∏ j = z k j d j ( k ) Z e Ω + g j ( t ) t k dV ( t ) ,where d j ( k ) = π n c k + ∑ jl = k l and g j ( t ) = f j ( √ t ) t k + ··· + k j − +( k j /2 ) . The assumption T φ m · · · T φ = A ( Ω ) then implies the equation(3) z k m ∏ j = z k j d j ( k ) Z e Ω + g j ( t ) t k dV ( t ) = z k ∈ A ( Ω ) . ERO PRODUCTS OF TOEPLITZ OPERATORS ON REINHARDT DOMAINS 7
Since Ω is bounded, we have A ( Ω ) ⊇ { z k } k ∈ N n . Moreover, there exists a multi-index k ∈ N n such that for any k ≥ k and any j the constant d j ( k ) >
0. Thus from (3) we have(4) m ∏ j = Z e Ω + g j ( t ) t k dV ( t ) = k ≥ k . Equivalently,(5) m ∏ j = Z e Ω + ( g j ( t ) t k + ) t k dV ( t ) = k ∈ N n . Set Z j = (cid:26) k ∈ N n : Z e Ω + ( g j ( t ) t k ) t k dV ( t ) = (cid:27) .Then (5) implies that ∪ mj = Z j = N n .Clearly N n satisfies condition (I). Then by Corollary 2, Z j satisfies condition (I) for some j .After a change of variables t = cx for sufficiently small c >
0, we may further assume that e Ω + ⊆ [
0, 1 ) n . Then the function h j ( z ) = Z e Ω + ( g j ( t ) t k ) t z dV ( t ) is bounded and holomorphic on H n + . Then Lemma 1 implies that h j ≡
0. Thus Z j = N n and, Stone-Weierstrass Theorem implies that g j ( t ) t k ≡
0. Therefore f j ≡ (cid:3) As a consequence of Proposition 1, we also have the following proof of Corollary 1.
Proof of Corollary 1.
Note that the bounded function φ m has an L expansion φ m ( r e i θ , . . . , r n e i θ n ) = ∑ p ∈ Z n f m , p ( r , . . . , r n ) e i p · θ where f m , p ( r , . . . , r n ) = ( π ) n Z π · · · Z π φ m ( r e i θ , . . . , r n e i θ n ) e − i p · θ d θ · · · d θ n are also bounded (see, for example, [Le10, Lemma 2.2] or [DZ11, Lemma 4.1]). From the as-sumption, the symbol functions φ , . . . , φ m − are bounded and quasi-homogeneous. Thus wecan set φ j ( z ) = f j ( r ) e i k j · θ for 1 ≤ j ≤ m − r = ( r , . . . , r n ) , θ j = ( θ , . . . , θ n ) , and k j ∈ Z n . A similar computation as in the proof of Proposition 1 then shows that for any z k with k ∈ N n , T φ m · · · T φ z k = ∑ p ∈ Z n z k + p d m , p ( k ) Z e Ω + g m , p ( t ) t k dV ( t ) m − ∏ j = z k j d j ( k ) Z e Ω + g j ( t ) t k dV ( t ) (6) = ˇZELJKO ˇCU ˇCKOVI ´C, ZHENGHUI HUO, AND S ¨ONMEZ S¸AHUTO ˘GLU where d m , p ( k ) = π n c k + p + ∑ m − l = k l , d j ( k ) = π n c k + ∑ jl = k l , g j ( t ) = f j ( √ t ) t k + ··· + k j − +( k j /2 ) , g m , p ( t ) = f m , p ( √ t ) t k + ··· + k m − +( p /2 ) .The first equality in (6) holds by the L convergence of the series ∑ p ∈ Z n f m , p ( r ) e i p · θ and theboundedness of z k . This together with the fact that k , . . . , k m − are fixed imply that for any k ∈ N n and p ∈ Z n we have z k + p d m , p ( k ) Z e Ω + g m , p ( t ) t k dV ( t ) m − ∏ j = z k j d j ( k ) Z e Ω + g j ( t ) t k dV ( t ) = φ j = j ∈
1, 2, . . . m − f m , p =
0. If φ j = j ∈
1, 2, . . . m −
1, then we are done. Otherwise, f m , p = p ∈ Z n which yields that φ m = (cid:3) Now we turn our attention to proving Theorem 1. We introduce the following notation whichwill be used in the proof. For multi-indices a = ( a , . . . , a n ) and b = ( b , . . . , b n ) we say a ≤ b if a j ≤ b j and a < b if a j < b j for all j . When a < b the boxR = { k ∈ Z n : a j ≤ k j < b j for j =
1, 2, . . . , n } is said to have dimension d = ( d , . . . , d n ) where d j = b j − a j for j =
1, 2, . . . , n . That is, d j is thesize of R in the j th component. Proof of Theorem 1.
Since φ j s are finite sums of bounded quasi-homogeneous functions, wemay assume that for j =
1, . . . , m φ j ( z ) = ∑ k ∈ R j φ j , k ( z ) .where φ j , k ( z ) = f j , k ( r ) e i k · θ are bounded quasi-homogeneous functions and R j is a box in Z n of dimensions d ( j ) = ( d ( j ) , . . . , d n ( j )) . We will prove the theorem using an induction on thedimensions of R , . . . , R m .Let l denote the multi-index in Z n whose l th entry is 1 and every other entry is 0. Whenall of R j s are of dimension at most (
1, . . . , 1 ) , the symbols φ j are bounded quasi-homogeneous.Then the statement of Theorem 1 follows by Proposition 1.Next, suppose the statement is proved when all of R j s have dimensions less than or equal to d ( j ) ≥ (
1, . . . , 1 ) . We claim the same result holds true when for some j , R j has dimensions d ′ ( j ) where d ′ ( j ) = d ( j ) + l while the dimensions of all other R j s are still less than or equalto d ( j ) ≥ (
1, . . . , 1 ) . ERO PRODUCTS OF TOEPLITZ OPERATORS ON REINHARDT DOMAINS 9
Without loss of generality, we may assume that l = φ j ( z ) = ∑ k ∈ R j φ j , k ( z ) , we have T φ m · · · T φ = ∑ k ∈ R · · · ∑ k m ∈ R m T φ m , k m · · · T φ k .Using the same notation as in the proof of Proposition 1, it follows that for z k ∈ A ( Ω ) ,0 = ∑ k ∈ R · · · ∑ k m ∈ R m T φ m , k m · · · T φ k z k = ∑ ( k ,..., k m ) ∈ R ×···× R m z k m ∏ j = z k j d j ( k ) Z e Ω + g j , k j ( t ) t k dV ( t ) .(7)Suppose R s = { α ∈ Z n : a sj ≤ α j < b sj + ≤ j ≤ n } for s = m . Then the dimensionsof R s s imply that b j − a j = d ′ ( ) − = d ( j ) and b s − a s = d ( s ) − s = j . Thusthe monomial z k ∏ mj = z k j contains the factor z k ∏ mj = z b j only if the first entries of k j are b j respectively (because α ≤ b j when α ∈ R j ). We set R j = { α ∈ R j : α = b j } ,and define φ j ( z ) = ∑ k ∈ R j φ j , k ( z ) .Then (7) implies that ∑ ( k ,..., k m ) ∈ R ×···× R m z k m ∏ j = z k j d j ( k ) Z e Ω + g j , k j ( t ) t k dV ( t ) = T φ m · · · T φ = A ( Ω ) , since the terms in (8) have the same factor z k ∏ mj = z b j ,the largest power of z in (7). Because R j has dimensions less than or equal to d ( j ) , the induc-tion hypothesis implies that φ j = j .If φ j =
0, then φ j ( z ) = ∑ p ∈ R j \ R j φ j , p ( z ) .Note that R j \ R j is a box of dimensions equal to d ( j ) and hence the induction hypothesis thenyields that φ j = j . On the other hand, if φ j = j = j , then the dimensionsof R j is reduced and φ j ( z ) = ∑ p ∈ R j \ R j φ j , p ( z ) where R j \ R j = { α ∈ R j : a j ≤ α < b j } . We set R j = { α ∈ R j : α = b j − } and define φ j ( z ) = ∑ k ∈ R j φ j , k ( z ) .Now we consider the equation T φ m · · · T φ j · · · T φ = A ( Ω ) . By the same argument, wehave either φ j = φ j = j = j . Again, if φ j =
0, then it follows, from theinduction hypothesis applied to T φ m · · · T φ , that φ j = j . If φ j = j = j , j then R j can be reduced to R j \ R j . If φ j =
0, then the dimension of R j will be further reduced.Repeating this process, we would either have φ j = R j s will have its dimensions decreasedby 1 along certain coordinate direction. Since each R j has finite dimensions, some R j will bereduced to an empty set after finitely many steps. This would imply that the correspondingsymbol φ j = (cid:3) R EFERENCES [A ˇC01] Patrick Ahern and ˇZeljko ˇCuˇckovi´c,
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